In Exercises 5–14, an objective function and a system of linear i...

Question

In Exercises 5–14, an objective function and a system of linear inequalities representing constraints are given.
a. Graph the system of inequalities representing the constraints. 
b. Find the value of the objective function at each corner of the graphed region. c. Use the values in part (b) to determine the maximum value of the objective function and the values of x and y for which the maximum occurs.

Accepted Answer

Welcome back. I am so glad you're here. We have quite a complex problem here. We are given a system of inequalities and were asked to graph it. That will be the first part. Then were asked to evaluate the objective function at each corner of the graft region. The objective function is this Z equals two X plus six Y. And then finally, we will be identifying the values for X and Y at which the maximum value is found. That will be the last part. So we're starting off with the graphing. I am going to draw a rough sketch of a coordinate plane here will have a vertical Y axis and a horizontal X axis. They come together at the origin in the middle, we have five inequalities here we are going to be graphing X is greater than or equal to zero. And then also this see the comma we have why is greater than or equal to zero? And I'm going to start with those two and then we'll move on to the other three. So for X is greater than or equal to zero. Where is that true? Where is X greater than, or equal to zero. Well, that is anywhere including the Y axis and to the right of the y axis. So it's heading to the right, where is it true that why is greater than, or equal to zero? Well, that is anywhere and including the X axis that is above the X axis, this direction, why did those disappear? And now where is it true that both X is greater than, or equal to zero? And why is greater than or equal to zero? That's in this region here where they're both shaded in. So that's in the first quadrant. So because we're only dealing with the first quadrant, I am going to erase this coordinate plane and I'm going to zoom in on the first quadrant. So really, we're just mostly dealing with where X and Y are both positive. So now kind of setting aside X is greater than zero and Y is greater than zero for a moment. Now dealing with those other three inequalities and keeping in mind that all five of these inequalities are either greater or less than, but also or equal to, they all include the or equal to. And so that means we're drawing solid lines for all five of these. So the X and the Y axis in these regions are included, but I'm not going to color those in yet. So now let's work on this negative three, X plus four Y is less than or equal to 20 and we recall from previous lessons when we're graphing inequalities, we first want to graph equality is we want to set them equal to. So we have negative three X plus four Y equals 20. And we want to start by graphing that in order to graph that, we can see that that's a line. So I'm going to set the X and Y intercepts, set X equal to zero, plug that in, find the Y value and set Y equal to zero, plug that in and find the X value. So plugging zero in for X, we have negative three times zero instead of X plus four, Y equals 20. So negative three times zero is zero. We have four, Y Equals 20 and dividing both sides by four, we get a y value of five graphing the 0. from the origin. We go neither left nor right for X value of zero. And we go up five units for our Y value of five and we label that 05. Now for our Y or for our Y value of zero, we'll figure out our X intercept. So we have negative three, X plus four and plugging in zero for Y equals 20. So four times +00, we have negative three, X equals 20 dividing both sides by negative three, we get X equals negative 20/ And negative 20/3 is way off of the chart of what we're dealing with somewhere over here. And so we can imagine continuing this line and then I'm going to draw it for the portion that we're dealing with. So it will continue in this direction. And now what we want to do because this is an inequality we want to figure out which region we're talking about. When we say that negative three X plus four, why is less than or equal to 20? Are we talking about this region up here to the left of it? Are we talking about the region down here? And to the right? And in order to discern that we choose test points, I am going to choose a test point at + For this region down here. And I'm going to choose a test point at 110 to figure out that region up there. So let's plug in for our X value one. We have negative three times now one, not X plus and then we'll test out 10 for our Y value. So plus four times 10, see if this is less than or equal to 20 negative three times one is negative three plus four times 10 is 40 is less than or equal to 20. Well, negative three plus 40 that's 37 that's not less than or equal to 20. We can reject this test point. And if we want, we don't need to for all of these, but we can show real quick to test 11. See that that one does work. So negative three times one for the X value plus four times one for the Y value is less than or equal to 20. Negative three times one is negative three plus four times one is four and negative three plus four is one and one is indeed less than 20. So we know we're talking about this region down into the, right. So let's update our graph. That's a lot of what this is just updating our graph. We're talking about this region down here now, but we've got two other inequalities to narrow this down a bit. So let's now work on four x plus y is less than, or equal to 24. And again, setting this to an equality, we have four X plus Y equals and we want to find our intercepts. So plugging in zero for X, we have four times zero plus Y equals 24. Well, four times 00. So we get a Y value of 24. Again, this is way off our charts. So 0 24 from the origin moving neither left nor right for our X value of zero And up 24 units for a y value of 24. I'm going to draw a dot up here just to let us know that that's the direction we're aiming for. And now putting zero in for the Y value, we have four times X plus not Y but zero equals 24 So this is just four X equals 24 dividing both sides by four, we get X equals six. So now plotting 60 from the origin, we're moving to the right six units for our X value of six and moving neither up nor down for our Y value of zero, we can label this 60 and then draw a line heading in the direction of that point that we had figured out. We know we're going that way. And now it is a matter of, are we talking about for our inequality for X plus Y is less than, or equal to 24? We're talking about this region to the right of our line or are we talking about this region to the left of our line? So choosing test points, I am going to choose again 11 for one test point and then I am going to choose another point at 10 1. Now let's plug in 11. So we have four times not X but one plus not why, but one is less than, or equal to 24 4 times one is four plus one and four plus one is five. Is it true that five is less than or equal to 24? Yes, it's true. So we're talking about that region containing the 240.11 where four X plus Y is less than or equal to 24. So now we can update our shaded region. We're talking about this region down here. And we just have one more inequality to go. I'm going to move this tables. We've got a little more room and now we will graph the inequality X plus two Y is greater than, or equal to four. And before we can figure out the inequality, we set it to an equality X plus two Y equals four. And we want to find the intercepts. So plugging in zero for X, we have zero plus two, Y equals four. Well, that's just two, Y equals four, dividing both sides by two. We get a Y value of two. Graphing the 0.2 from the origin going neither left nor right for our X value of zero and up two units for A Y value of two and labeling this 02. Now figuring out where our X intercept is so plugging in zero for Y, we have X plus two times zero equals +42 times zero is zero. So we have X equals four. So now graphing the 0.40 from the origin, we're moving to the right four units for X value of four, moving neither up nor down for our Y value of zero. Labeling this 40. Now we can connect these two points. It's a line, it would keep going. But this is the region that we're interested in. Now, we want to know where are we talking about? Are we talking about the region up here? To the right of the line or are we talking about the region down into the left of this line for our inequality? And so we can choose test points. I am going to start with the test point of 00 for this down into the left. And so four X that will be zero plus two times our Y value is zero is greater than, or equal to four. Well, the left hand side is just zero plus two times 00. That's zero. Is it true that zero is greater than or equal to four? Nope, it is not. So we can reject this area down into the left where our test point of 00 is and we can refine our shading even more. It's going to be this area up and to the right of that latest line, which is this area here. And now recall that we also included the X and Y axis. So we can connect these points 05 to 0 to and we can connect to the points 40 in 60. And look there is our beautiful graph except one thing we need to know the points for all of the edges. And we're missing this one. We have to figure out where those two lines come together. So again, we're going to set up a smaller this time system of equations. It's going to be equations not inequalities. We want to figure out where our line negative three, X plus four, Y equals 20. And the line four X plus Y equals 24 come together, we want to know those X and Y values. So let's take a little detour down here. And now recalling previous lessons from solving systems of equations, we can use the addition method, we can add these together. I am going to adjust this second Equation and I'm going to multiply everything times a -4. So we have four, X times negative four will be negative 16 X. Why times negative four will be minus four, Y equals 24 times negative four will be negative 96. So now adding, I'll label these the first equation and the new second equation here adding those two together, we will have negative three X plus negative 19 X gives us an excuse me, negative three X plus negative 16 X gives us negative 19 X. The Y values cancel out as designed four, Y minus four, Y is zero wise. So now we can solve for X and 20 minus 96 equals a negative 76. So we have negative 19 X equals a negative 76. We can divide both sides by negative 19. And we solve for X, we get an X value of four. Now we can plug that X value back in to either of the equations. Let's choose the original second equation. So we'll have four times no longer X but now four plus Y equals 24 that simplifies to 16 plus Y equals 24 4 times 16. We can subtract the 16 from both sides of the equation and we get a Y equal to 24 minus 16 is eight. So we've solved for our X and Y values where those two lines come together. So I'm going to write 48 and then scroll back up and we can label that point on our graph that we have created. Oh, my goodness. Alright, sorry for that. This is the .48. Alright. We've got all of our corners identified. So that's that's step one of graphing the system of inequalities. Now we get to do that part where we evaluate the objective function. So how do we do that? We are going to take the X and Y values for each of our corners and we're going to plug those in for X and Y in that Z equals two X plus six Y objective function. And now we've got a little bit more room to work and we can see everything that we need to see. Let's do that. So let's start with the point of 02. So plugging in we have two times no longer X but now zero for our X value plus six times no longer. Why? But now two for our Y value two times 00 and six times two is 12. So this equals 12. Now let's evaluate for the corner at 05. So we have two times x value is zero Plus six times. The y value is five, two times 006 times five is 30. So this equals 30 0 plus 30 is 30. Now let's look at the .4, 8, two times the X value of four plus six times. The Y value of eight, two times four is eight plus six times eight is 48 plus 48 is 56. Now let's evaluate 460, two times. The X value is six plus six times. The Y value is zero and two times six is 12 plus six times +00. So this is just 12. Now, the last one at 40, two times the X value is four plus six times. The Y value is 02 times four is eight plus six times zero is zero. So eight plus zero is eight. Okay. The last step from our directions is to identify the maximum value. The maximum value is the highest one of these five corners and the highest one is where talking about the corner 48 where it's equal to 56. Now, we just need to find the appropriate answer choice. So we have got corners at 02054860 and 40 answer choice. A has all of those we want for our Z value at zero to that equals 12. We've got that at 05, Z is equal to 30. We've got that at 48, Z is equal to 56. We've got that at 60, Z is equal to 12. We've got that at 40, Z is equal to eight. We've got that. The maximum value is 56 at X equals four and Y equals eight. We've got that. It's been right in front of us all along. Answer choice a well done. We'll catch you on the next one.

Key Concepts

Linear Inequalities

Objective Function

Corner Point Theorem

Watch next